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Ch.14 - Chemical Kinetics

Chapter 14, Problem 76a

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important?

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Hi everyone for this problem. We're told the particle size of a heterogeneous catalyst is essential to its function. Which of the following is the most efficient heterogeneous catalyst. So, something important here is that we're told the particle size is essential to its function and we want to know which of the following answer choices is going to be the most efficient. So because we're looking at particle size, let's look at it in terms of what's going to increase the rate of our reaction. So if we have a small particle size, what that means is we're going to have a larger surface area. Okay. And by having a larger surface area. This is important when it comes to active sites, by having a larger surface area, we have more active sites, which is going to increase the reaction rate. So for all of the answers, choices given our smart smallest particle size is going to be the most efficient and fine powder is going to be our smallest particle size. So this is going to make answer choice B our most efficient heterogeneous catalyst. Okay. Due to that particle size, that's the end of this problem. And that's the answer to this problem. I hope this was helpful
Related Practice
Textbook Question

You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g)

You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism:

HBr(g) + O2(g) → HOOBr(g)

HOOBr(g) + HBr(g) → 2 HOBr(g)

HOBr(g) + HBr(g) → H2O(g) + Br2(g)

(a) Confirm that the elementary reactions add to give the overall reaction.

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Textbook Question

You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g)

You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism:

HBr(g) + O2(g) → HOOBr(g)

HOOBr(g) + HBr(g) → 2 HOBr(g)

HOBr(g) + HBr(g) → H2O(g) + Br2(g)

(b) Based on the experimentally determined rate law, which step is rate determining?

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Textbook Question

(c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

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Textbook Question

The addition of NO accelerates the decomposition of N2O, possibly by the following mechanism: NO1g2 + N2O1g2¡N21g2 + NO21g2 2 NO21g2¡2 NO1g2 + O21g2 (b) Is NO serving as a catalyst or an intermediate in this reaction?

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Textbook Question

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina 1Al2O32 or silica 1SiO22. (b) How does the surface area affect the rate of reaction?

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Textbook Question

The enzyme urease catalyzes the reaction of urea, 1NH2CONH22, with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of 4.15 * 10-5 s-1 at 100 C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 * 104 s-1 at 21 C. (c) In actuality, what would you expect for the rate of the catalyzed reaction at 100 C as compared to that at 21 C?

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